The net force of an object moving down a incline with coefficient of kinetic frictio

A crate is on a 20 degree plane where the coefficients of static and kinetic friction are .45 and .35 respectively. What is the acceleration of the crate?

2. Relevant equations

Mk= .35
Ms= .45
F= sin20 * mg (I think this is the net force?)

3. The attempt at a solution

F - Fk= M*A

(sin20 * mg) - (.35 * mg) = m * a

a = sin20 * g - .35 * g

a= .34 * 9.8 - .35 * 9.8

a = -.098

The answer in the book says 0. As I was writing this problem I think I figured it out. If sin 20 is the net force, then that would make net force less than the static friction which would give it 0 acceleration. Maybe I just miscalculated the net force. If so could someone explain to me what net force would be for this equation? Thanks, guys.